    Question

# The equation of the circle passing through points of intersection of the circle x2+y2−2x−4y+4=0 and the line x+2y=4 and touches the line x+2y=0, is

A
x2+y2+x+2y=0
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B
x2+y2+x2y=0
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C
x2+y2x2y=0
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D
x2+y2x+2y=0
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Solution

## The correct option is C x2+y2−x−2y=0Equation of any circle passing through points of intersection of the given circle and the line is (x2+y2−2x−4y+4)+λ(x+2y−4)=0 ⇒x2+y2+(λ−2)x+(2λ−4)y+4(1−λ)=0⋯(1) It will touch the line x+2y=0 if solution of equation (1) and x+2y=0 is unique. Hence the roots of the equation (−2y)2+y2+(λ−2)(−2y)+(2λ−4)y+4(1−λ)=0 ⇒5y2+4(1−λ)=0 must be equal. 0−4×5×4(1−λ)=0 ⇒λ=1 From (1), the required circle is x2+y2−x−2y=0  Suggest Corrections  0      Similar questions  Related Videos   Circle and Point on the Plane
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